Let `f :R->R` be a function defined by `f(x)=|x]` for all `x in R` and let `A = [0, 1),` then `f^-1 (A)` equals

Options

A. (-1,1)
B. (0,1)
C. (-1,0)
D. None of these

Explanation

To solve the problem, we need to find the inverse image of the set \( A = [0, 1) \) under the function \( f(x) = |x| \).

### Step-by-Step Solution:

1. **Understand the function and the set**:
The function \( f(x) = |x| \) takes any real number \( x \) and returns its absolute value. The set \( A = [0, 1) \) includes all real numbers \( y \) such that \( 0 \leq y < 1 \).

2. **Set up the condition for the inverse image**:
We want to find all \( x \in \mathbb{R} \) such that \( f(x) \in A \). This means we need to find \( x \) such that:
\[ f(x) = |x| \in [0, 1) \]

3. **Translate the condition**:
The condition \( |x| < 1 \) implies:
\[ -1 < x < 1 \]
This is because the absolute value \( |x| \) is less than 1 when \( x \) is in the open interval \( (-1, 1) \).

4. **Consider the non-negativity of the absolute value**:
Since \( |x| \geq 0 \) for all \( x \), we also have the condition that \( |x| \) must be greater than or equal to 0, which is always satisfied for any real number \( x \).

5. **Combine the conditions**:
Therefore, the only restriction we have is from the condition \( |x| < 1 \), leading us to conclude:
\[ f^{-1}(A) = (-1, 1) \]

### Final Result:
Thus, the inverse image \( f^{-1}(A) \) is:
\[ f^{-1}(A) = (-1, 1) \]